Is the following a covering space of $S^1 \vee S^1$ ?
It would appear so since there is no point that has more than 2 incoming or outgoing arrows. It seems that the potential covering map $p:Y\to S^1 \vee S^1$ fulfils all the requirements, however I'm not completely sure.
Please help me understand this better.
It is indeed a covering space.
You can see this with some measure of rigor by using the arrows in the diagram as a way to orient each edge of $Y$ (as you have already noticed), but on top of that you need to convince yourself that it is possible to assign the letters $a,b$ to the edges of $Y$ so that at each vertex of $Y$ there is exactly one incoming $a$ edge, exactly one outgoing $a$ edge, exactly one incoming $b$ edge, and exactly one outgoing $b$ edge. Next you also label $S^1 \vee S^1$ by labelling one of the $S^1$'s as the $a$ loop, and labelling the other $S^1$ as the $b$ loop. Then you can define a covering map $Y \mapsto S^1 \vee S^1$: map each $a$ edge of $Y$ around the $a$ loop of $S^1 \vee S^1$ preserving orientation, and map each $b$ edge of $Y$ around the $b$ loop of $S^1 \vee S^1$ preserving orientation. Check that this defines a covering map from $Y$ to $S^1 \vee S^1$ and you are done.
If you wish to have a still higher level of rigor, you will need to come up with a more rigorous description of $Y$ to begin with.